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Discrete Structures Chapter 5 Relations

Discrete Structures Chapter 5 Relations. Nurul Amelina Nasharuddin Multimedia Department. Objectives. On completion of this topic, student should be able to: Determine the properties of relations – reflexive, symmetric, transitive, and antisymmetric

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Discrete Structures Chapter 5 Relations

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  1. Discrete StructuresChapter 5 Relations Nurul Amelina Nasharuddin Multimedia Department

  2. Objectives • On completion of this topic, student should be able to: • Determine the properties of relations – reflexive, symmetric, transitive, and antisymmetric • Determine equivalence and partial order relations • Represent relations using matrix and graph

  3. Outline • Properties of relations • Matrix and graph representation of relations • Equivalence relations • Partial order relations

  4. Recall: Relations • The most basic relation is “=” (e.g. x = y) • Generally x R y  TRUE or FALSE • R(x,y) is a more generic representation • R is a binary relation between elements of some set A to some set B, where xA and yB • Binary relations: x R y On sets xX, yY R  X  Y

  5. Example • “less than” relation from A={0,1,2} to B={1,2,3} • Use traditional notation • 0 < 1, 0 < 2, 0 < 3, 1 < 2, 1 < 3, 2 < 3 • Or use set notation • AB={(0,1),(0,2),(0,3),(1,1),(1,2),(1,3),(2,1),(2,2), (2,3)} • R={(0,1),(0,2),(0,3),(1,2),(1,3),(2,3)}  AB • Or use Arrow Diagrams

  6. Properties of Relations • Reflexive: Let R be a binary relation on a set A • Eg: Let A = {1,2,3,4} • R1 ={(1,1),(1,2),(2,2),(2,3),(3,3),(4,4)} • R1 is reflexive • R2 = {(1,1),(2,2),(3,3)} •  R2 is not reflexive (why?)

  7. Properties of Relations • Symmetric: Let R be a binary relation on a set A • Eg: Let A = {1,2,3} • R1 ={(1,2),(2,1),(1,3),(3,1)} • R1 is symmetric • R2 = {(1,1),(2,2),(3,3),(2,3)}. •  R2 is not symmetric because (3,2) R2

  8. Properties of Relations • Transitive: Let R be a binary relation on a set A • Let A = {1,2,3,4} • R = {(2,1),(3,1),(3,2),(4,1),(4,2),(4,3)} • R is transitive because • (3,2) & (2,1) → (3,1) • (4,2) & (2,1) → (4,1) • (4,3) & (3,1) → (4,1) • (4,3) & (3,2) → (4,2)

  9. Properties of Relations xRy or (x,y)  R for all x and y in A, R is reflexive  for all x in A, (x,x)  R R is symmetric  for all x and y in A, if (x,y)  R then (y,x)  R R is transitive  for all x, y and z in A, if (x,y)  R and (y,z)  R then (x,z)  R

  10. Example (1) • Define a relation of A called R • A = {2,3,4,5,6,7,8,9} • R = {(4,4),(4,7),(7,4),(7,7),(2,2),(3,3),(3,6),(3,9), (6,6),(6,3),(6,9),(9,9),(9,3),(9,6)} • Is R • Reflexive? No • Symmetric? Yes • Transitive? Yes

  11. Example (2) • A = {0,1,2,3} • R over A = {(0,0),(0,1),(0,3),(1,0), (1,1),(2,2),(3,0),(3,3)} • Is R • Reflexive? Yes. • Symmetric? Yes. • Transitive? No. (1,0),(0,3)  R but (1,3)  R

  12. Proving Properties on Infinite Sets-“equal” relation (1) • Define a relation R on R (the set of all real numbers): • For all x, y  R, x R y ↔ x = y • Is R reflexive? symmetric? transitive? • R is reflexive iffx R, x R x. By definition of R, this means x = x, for all x R. This is true, since every real number is equal to itself. Hence, R is reflexive

  13. Proving Properties on Infinite Sets-“equal” relation (1) • R is symmetric iff x,y R, if x R y then y R x. By definition of R, this means that x,y R, if x = y then y = x. So this is true; if one number is equal to a second, then the second is equal to the first. Hence, R is symmetric • R is transitive iff x,y R, if x = y and y = z, then x = z. So this is true; if one real number equals to a second, and the second equals a third, then the first equals the third

  14. Proving Properties on Infinite Sets -“less than” relation (1) • Define a relation R on R (the set of all real numbers): • For all x, y  R, x R y ↔ x < y • Is R reflexive? symmetric? transitive?

  15. Proving Properties on Infinite Sets -“less than” relation (2) • R is reflexive iff x R, x R x. By definition of R, this means x < x, for all x R. But this is false. Hence, R is not reflexive • R is symmetric iff x,y R, if x R y then y R x. By definition of R. This means that x,y R, if x < y then y > x. But this is false. Hence, R is not symmetric • R is transitive iff x,y,z R, if x < y and y < z, then x < z. Hence, R is transitive

  16. Properties of Congruence Modulo 3 (1) • Define a relation R on Z: For all m,n  Z, • m R n  3|(m – n) • R is called congruence modulo 3 • Is R reflexive? • Is R symmetric? • Is R transitive?

  17. Properties of Congruence Modulo 3 (2) • R is reflexive iff for all m in Z, m R m. By definition of R, this means 3|m – m or 3|0. This is true since 0 = 0 . 3, so R is reflexive • R is symmetric iff for all m,n in Z, m R n then n R m. By definition of R, this means if 3|(m – n) then 3|(n – m). This is true • m – n = 3k, for some integer k • n – m = - (m – n) = 3(-k) • Hence 3|(n – m). Therefore R is symmetric

  18. Properties of Congruence Modulo 3 (2) • R transitive iff for all m,n,p Z, if m R n and n R p then m R p. By definition of R, if 3|(m – n) and 3|(n – p) then 3|(m – p). So by definition of divide • m – n = 3r for some r • n – p = 3s for some s • It is crucial to observe that (m – n) + (n – p) = m – p • (m – n) + (n – p) = m – p = 3r + 3s • m – p = 3(r + s) • Hence 3|(m – p). Therefore, R is transitive

  19. Matrix Representation of a Relation • MR = [mij] (where i=row, j=col) • mij={1 iff (i,j)  R and 0 iff (i,j)  R} • Eg: R : {1,2,3}{1,2} where x > y • R = {(2,1),(3,1),(3,2)}

  20. Example • Example: • A = {1,2,3,4}, B = {w,x,y,z} • R = {(1,x),(2,x),(3,y),(3,z)} = zero-one matrix

  21. Graph Representation of a Relation • Let A = {1,2,3,4} and R is relation on A where R = {(1,1),(1,2),(2,3),(3,2),(3,3),(3,4),(4,2)}

  22. Union, Intersection, Difference and Composition of Relations • R: AB and S: AB • R: AB and S: BC

  23. Composition of Relations • Example: A = {1,2,3,4}, B = {w,x,y,z}, C = {5,6,7}, • R1: AB = {(1,x),(2,x),(3,y),(3,z)} and • R2: BC = {(w,5),(x,6)} • Therefore, R2o R1 = {(1,6),(2,6)}

  24. Equivalence Relations • Any binary relation that is: • Reflexive • Symmetric • Transitive • Eg: A = {1,2,3}, R = {(1,1),(2,2),(2,3),(3,2),(3,3)} • R is reflexive, symmetric and transitive • Therefore, R is an equivalence relation • Recall example earlier: Congruence modulo 3 is an equivalence

  25. Antisymmetry Relations • Let R be a relation on a set A. R is antisymmetric iff for all a and b in A, if a R b and b R a then a = b • In other words, a relation is antisymmetric iff there are no pairs of distinct elements a and b with a related to b and b related to a • Eg: Let A = {0,1,2} • R1= {(0,2),(1,2),(2,0)}. R1 is not antisymmetric • R2 = {(0,0),(0,1),(0,2),(1,1),(1,2)}. R2 is antisymmetric

  26. Example (1) • Let A = {1,2,3}, R = {(1,2),(2,1),(2,3)} • R is not symmetric, (3,2)  R • R is also not antisymmetric because (1,2),(2,1)  R • Let A = {1,2,3}, S = {(1,1),(2,2)} • S is both symmetric and antisymmetric

  27. Example (2) • Let R1 be the divides relation (a|b) on Z+ • Is R1 antisymmetric? Prove or give counterexample • If a R1 b and b R1 a, then a = b • a R1 b means a|b → b = k1a • b R1 a means b|a → a = k2b • It follows that, b = k1a = k1(k2b) = (k1k2)b • Thus, k1 = k2 = 1. Hence a = b • R1 antisymmetric

  28. Example (2) • Let R2 be the divides relation on Z • Is R2 antisymmetric? Prove or give counterexample. • Let a = 2, b = -2 • Hence, a|b (-2 = -1(2)) and • b|a (2 = -1(-2)) but a  b • R2 is not antisymmetric

  29. Partial Order Relations • Any binary relation that is: • Reflexive • Antisymmetric • Transitive • Partial Order Set (POSET) • (S,R) = R is a partial order relation on set S • Examples: • (Z, ) • (Z+,|) {note: | symbolizes divides}

  30. Quiz 5 • Exercise 10.2 (No. 1, 2, 16) • Exercise 10.5 (No. 2, 5) Send your answers in the next class!

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